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charle [14.2K]
3 years ago
10

Platinum has a work function (also called binding energy) of 9.05 × 10-19 J. Which is the longest wavelength that could cause em

ission of electrons? 2.196 × 106 m 5.654 × 102 m 2.196 × 10-7 m 1.37 × 1015 m 4.553 × 10-6 m
Physics
1 answer:
son4ous [18]3 years ago
6 0

Answer:

2,196 10⁻⁷ m

Explanation:

This is a problem of photoelectric effect, which was explained by Einstein, assuming that the light was formed by quanta of energy that collide with the electors and is described by the expression

         K_{max} = h f -Ф

Where   K_{max} is the maximum kinetic energy of the expelled electrons h is the Planck constant that is worth 6,626 10-34 J s, f is the radiation frequency and Ф is the material's work function.

For the wavelength of greater wavelength (lower energy) the kinetic energy of the electors must decrease to a minimum of zero.

         0 = hf -Ф

         hf = Ф

         f = Ф / h

         f = 9.05 10-19 / 6,626 10-34

         f = 1,366 10 15 Hz

Now the speed of the wave is related to the wavelength and frequency

        c = λ f

        λ = c / f

        λ = 3 10⁸ / 1,366 10¹⁵

        λ = 2,196 10⁻⁷ m

The result is 2,196 10⁻⁷ m

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
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Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

<h3>Step B:</h3>

<em>Using the same method as above:</em>

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

<em>Finally we could calculate the total work done and heat added as follows:</em>

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

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